A particle of mass 4 kg is connected to another particle of mass 5 kg by a taut light inelastic string which passes over a light smooth pulley at the edge of a rough horizontal table - Leaving Cert Applied Maths - Question 4 - 2015

Applying Newton's second law:

For the 5 kg mass: [ 5g - T = 5a \quad (1) ]

For the 4 kg mass: [ T - 0.75R = 4a \quad (2) ]

Since the 4 kg mass is on a rough surface, we have: [ R = 4g \quad (3) ] [ F = 0.75(4g) = 3g \quad (4) ]

Using equation (3) in (2) and substituting: [ T = 0.75(4g) + 4a ] Substituting T from equation (1): [ 5g - 5a = 0.75(4g) + 4a ] [ 5g - 3g = 9a ] [ a = \frac{2g}{9} = 2.22 \text{ m/s}^2 ]

Substituting the value of acceleration into equation (1): [ T = 5g - 5a \quad (5) ] [ T = 5(9.8) - 5(2.22) ] [ T = 49 - 11.1 = 37.9 N ]

For the 12 kg mass: [ T - 12g = -12a \quad (1) ]

For the 10 kg mass on the inclined plane: [ 10g \sin \alpha - T = 10a \quad (2) ]

With (\tan \alpha = \frac{3}{5}), we have: [ \sin \alpha = \frac{3}{\sqrt{34}} \quad (3) ] [ T = 10g \sin \alpha + 10a \quad (4) ]

Substituting back: From equation (1): [ T = 12g + 12a ] Setting equations (3) and (4) equal yields: [ 12g + 12a = 10g \frac{3}{\sqrt{34}} + 10a ] Solving for a results in: [ a = 1.82 \text{ m/s}^2 ]

Using the common acceleration in equation (1): [ T = 12g + 12a ] [ T = 12(9.8) - 12(1.82) ] [ T = 117.6 - 21.84 = 95.76 N ]